A, B and C can do a piece of work in 15 days, 20 days and 25 days respectively. They work one at a time, one day each in turn: A on the first day, B on the second day, C on the third day, A again on the fourth day, and so on. In how many days will the work be finished?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(18\frac{9}{10}\)
Take the work as the LCM of 15, 20 and 25, that is 300 units. Count full 3-day rounds first, then see who finishes the little that is left.
Daily work: A \(=\frac{300}{15}=20\) units, B \(=\frac{300}{20}=15\) units, C \(=\frac{300}{25}=12\) units
One round of 3 days (A, B, C) \(=20+15+12=47\) units
6 rounds \(=6\times47=282\) units in \(6\times3=18\) days
Work left \(=300-282=18\) units, done by A on day 19
A needs \(\frac{18}{20}=\frac{9}{10}\) of a day for 18 units
Total time \(\displaystyle =18\frac{9}{10}\) days